Question: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}6x-3y &= 1 \\ 7x-6y &= 2\end{align*}$
Answer: Begin by moving the $x$ -term in the second equation to the right side of the equation. $-6y = -7x+2$ Divide both sides by $-6$ to isolate $y$ $y = {\dfrac{7}{6}x - \dfrac{1}{3}}$ Substitute this expression for $y$ in the first equation. $6x-3({\dfrac{7}{6}x - \dfrac{1}{3}}) = 1$ $6x - \dfrac{7}{2}x + 1 = 1$ Simplify by combining terms, then solve for $x$ $\dfrac{5}{2}x + 1 = 1$ $\dfrac{5}{2}x = 0$ $x = 0$ Substitute $0$ for $x$ back into the top equation. $6( 0)-3y = 1$ $-3y = 1$ $-3y = 1$ $y = -\dfrac{1}{3}$ The solution is $\enspace x = 0, \enspace y = -\dfrac{1}{3}$.